LAGRANGE POLYNOMIAL

In numerical analysis, a 'Lagrange polynomial', named after Joseph Louis Lagrange, is the interpolation polynomial for a given set of data points in the 'Lagrange form'. It was first discovered by Edward Waring in 1779 and later rediscovered by Leonhard Euler in 1783.
As there is only one interpolation polynomial for a given set of data points it is a bit misleading to call the polynomial the Lagrange interpolation polynomial. The more precise name is interpolation polynomial in the Lagrange form.

Contents

Definition

Proof

Main idea

Usage

Example

Notes

See also

External links

Definition

Given a set of ''k'' + 1 data points
:$(x\_0,\; y\_0),ldots,(x\_k,\; y\_k)$
where no two ''x''_''j'' are the same, the 'interpolation polynomial in the Lagrange form' is a linear combination
:$L(x)\; :=\; sum\_\{j=0\}^\{k\}\; y\_j\; ell\_j(x)$
of Lagrange basis polynomials
:$ell\_j(x)\; :=\; prod\_\{i=0,,\; i$
eq j}^{k} rac{x-x_i}{x_j-x_i} = rac{x-x_0}{x_j-x_0} cdots rac{x-x_{j-1}}{x_j-x_{j-1}} rac{x-x_{j+1}}{x_j-x_{j+1}} cdots rac{x-x_{k}}{x_j-x_{k}}.

Proof

The function we are looking for has to be a polynomial function ''L''(''x'') of degree less than or equal to ''k'' with
:$L(x\_j)\; =\; y\_j\; qquad\; j=0,ldots,k$
The Lagrange polynomial is a solution to the interpolation problem:
As can be easily seen
# $ell\_j(x)$ is a polynomial and has degree ''k''.
# $ell\_i(x\_j)\; =\; delta\_\{ij\},quad\; 0\; leq\; i,j\; leq\; k.,$
Thus the function ''L''(''x'') is a polynomial with degree at most ''k'' and
:$L(x\_i)\; =\; sum\_\{j=0\}^\{k\}\; y\_j\; ell\_j(x\_i)\; =\; y\_i.$
There can be only one solution to the interpolation problem since the difference of two such solutions would be a polynomial with degree at most ''k'' and ''k+1'' zeros.
Therefore ''L''(''x'') is our unique interpolation polynomial.

Main idea

Solving an interpolation problem leads to a problem in linear algebra where we have to solve a matrix. Using a standard monomial basis for our interpolation polynomial we get the very complicated Vandermonde matrix. By choosing another basis, the Lagrange basis, we get the much simpler identity matrix = δ_''i'',''j'' which we can solve instantly.

Usage

Example

We wish to interpolate $f(x)=\; an(x)$ at the points

$x\_0=-1.5$	$f(x\_0)=-14.1014$
$x\_1=-0.75$	$f(x\_1)=-0.931596$
$x\_2=0$	$f(x\_2)=0$
$x\_3=0.75$	$f(x\_3)=0.931596$
$x\_4=1.5$	$f(x\_4)=14.1014$

The basis polynomials are:
:$ell\_0(x)=\{x\; -\; x\_1\; over\; x\_0\; -\; x\_1\}cdot\{x\; -\; x\_2\; over\; x\_0\; -\; x\_2\}cdot\{x\; -\; x\_3\; over\; x\_0\; -\; x\_3\}cdot\{x\; -\; x\_4\; over\; x\_0\; -\; x\_4\}$
={1over 243} x (2x-3)(4x-3)(4x+3)

:$ell\_1(x)=\{x\; -\; x\_0\; over\; x\_1\; -\; x\_0\}cdot\{x\; -\; x\_2\; over\; x\_1\; -\; x\_2\}cdot\{x\; -\; x\_3\; over\; x\_1\; -\; x\_3\}cdot\{x\; -\; x\_4\; over\; x\_1\; -\; x\_4\}$
=-{8over 243} x (2x-3)(2x+3)(4x-3)

:$ell\_2(x)=\{x\; -\; x\_0\; over\; x\_2\; -\; x\_0\}cdot\{x\; -\; x\_1\; over\; x\_2\; -\; x\_1\}cdot\{x\; -\; x\_3\; over\; x\_2\; -\; x\_3\}cdot\{x\; -\; x\_4\; over\; x\_2\; -\; x\_4\}$
={1over 243} (243-540x^2+192x^4)

:$ell\_3(x)=\{x\; -\; x\_0\; over\; x\_3\; -\; x\_0\}cdot\{x\; -\; x\_1\; over\; x\_3\; -\; x\_1\}cdot\{x\; -\; x\_2\; over\; x\_3\; -\; x\_2\}cdot\{x\; -\; x\_4\; over\; x\_3\; -\; x\_4\}$
=-{8over 243} x (2x-3)(2x+3)(4x+3)

:$ell\_4(x)=\{x\; -\; x\_0\; over\; x\_4\; -\; x\_0\}cdot\{x\; -\; x\_1\; over\; x\_4\; -\; x\_1\}cdot\{x\; -\; x\_2\; over\; x\_4\; -\; x\_2\}cdot\{x\; -\; x\_3\; over\; x\_4\; -\; x\_3\}$
={1over 243} x (2x+3)(4x-3)(4x+3)

Thus the interpolating polynomial then is
:$\{1over\; 243\}Big(f(x\_0)x\; (2x-3)(4x-3)(4x+3)-8f(x\_1)x\; (2x-3)(2x+3)(4x-3)$
:::$+f(x\_2)(243-540x^2+192x^4)-8f(x\_3)x\; (2x-3)(2x+3)(4x+3)\; ,$
:::$+f(x\_4)x\; (2x+3)(4x-3)(4x+3)Big),$
:$=-1.47748x+4.83456x^3.,$

Notes

The Lagrange form of the interpolation polynomial shows the linear character of polynomial interpolation and the uniqueness of the interpolation polynomial. Therefore, it is preferred in proofs and theoretical arguments. But, as can be seen from the construction, each time a node ''x''_''k'' changes, all Lagrange basis polynomials have to be recalculated. A better form of the interpolation polynomial for practical (or computational) purposes is the Newton polynomial. Using nested multiplication amounts to the same idea.
Lagrange and other interpolation at equally spaced points, as in the example above, yield a polynomial oscillating above and below the true function. This oscillation is lessened by choosing interpolation points at Chebyshev nodes.
The Lagrange basis polynomials are used in numerical integration to derive the Newton-Cotes formulas.
Lagrange interpolation is often used in digital signal processing of audio for the implementation of fractional delay FIR filters (e.g., to precisely tune digital waveguides in physical modelling synthesis).

See also

★ Polynomial interpolation

★ Newton form of the interpolation polynomial

★ Bernstein form of the interpolation polynomial

★ Newton-Cotes formulas

★ Lagrange Method of Interpolation - Notes, PPT, Mathcad, Mathematica, Matlab, Maple at Holistic Numerical Methods Institute

External links

★ Lagrange interpolation polynomial on www.math-linux.com

★

★ Module for Lagrange Polynomials by John H. Mathews